Let $f(x)=x^3+3x^2-9x+7$. On which intervals is $f$ increasing? Choose 1 answer: Choose 1 answer: (Choice A) A $-3<x<1$ only (Choice B) B $x>1$ only (Choice C) C $x<-3$ and $x>1$ (Choice D) D $x<1$ only (Choice E) E The entire domain of $f$
Answer: We can analyze the intervals where $f$ is increasing/decreasing by looking for the intervals where its derivative $f'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. The derivative of $f$ is $f'(x)=3(x+3)(x-1)$. $f'(x)=0$ for $x=-3,1$. Since $f'$ is a polynomial, it's defined for all real numbers. Therefore, our critical points are $x=-3$ and $x=1$. $f$ is defined for all real numbers so we only need to consider the critical points. Our critical points divide the number line into three intervals: $\llap{-}7$ $\llap{-}5$ $\llap{-}3$ $\llap{-}1$ $1$ $3$ $5$ $7$ $\begin{array}{rl} x<& \llap{-}3\end{array}$ $ \llap{-}3<x<1$ $ x>1$ Let's evaluate $f'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $f'(x)$ Verdict $x<-3$ $x=-4$ $f'(-4)=15>0$ $f$ is increasing. $\nearrow$ $-3<x<1$ $x=0$ $f'(0)=-9<0$ $f$ is decreasing. $\searrow$ $x>1$ $x=2$ $f'(2)=15>0$ $f$ is increasing. $\nearrow$ In conclusion, $f$ is increasing over the intervals $x<-3$ and $x>1$.